The speed of a body moving with uniform acceleration is u. This speed is doubled while covering a distance S. When it covers an additional distance S, its speed would become

Options

A. `sqrt3u`
B. `sqrt5u`
C. `sqrt11u`
D. `sqrt7u`

Explanation

To solve the problem step by step, we will use the equations of motion under uniform acceleration.

### Step 1: Understand the initial conditions
The body starts with an initial speed \( u \) and moves with uniform acceleration. After covering a distance \( S \), its speed doubles.

### Step 2: Apply the third equation of motion
The third equation of motion states:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( s \) is the distance covered.

### Step 3: Substitute the known values after the first distance \( S \)
After covering the distance \( S \), the speed becomes \( 2u \). Therefore, we can substitute \( v = 2u \) and \( s = S \) into the equation:
\[ (2u)^2 = u^2 + 2aS \]
This simplifies to:
\[ 4u^2 = u^2 + 2aS \]
Rearranging gives:
\[ 4u^2 - u^2 = 2aS \implies 3u^2 = 2aS \]
Thus, we can express \( 2aS \) as:
\[ 2aS = 3u^2 \]

### Step 4: Find the speed after covering an additional distance \( S \)
Now, we need to find the speed after covering an additional distance \( S \). The total distance covered now is \( S + S = 2S \). The initial speed for this segment is \( 2u \).

Using the third equation of motion again:
\[ v^2 = (2u)^2 + 2a(2S) \]
Substituting \( 2u \) for the initial speed and \( 2S \) for the distance:
\[ v^2 = 4u^2 + 2a(2S) = 4u^2 + 4aS \]
Now, we can substitute \( 2aS \) from our earlier result:
\[ v^2 = 4u^2 + 4 \left(\frac{3u^2}{2}\right) \]
This simplifies to:
\[ v^2 = 4u^2 + 6u^2 = 10u^2 \]

### Step 5: Solve for \( v \)
Taking the square root gives:
\[ v = \sqrt{10}u \]

### Final Answer
The speed after covering an additional distance \( S \) is \( \sqrt{10}u \).

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